The value of c in Rolle's theorem for the function

Solution:

(C) $\frac{1-\sqrt{5}}{2}$

Given:

$f(x)=\frac{x(x+1)}{e^{x}}$

Differentiating the given function with respect to x, we get 

$f^{\prime}(x)=\frac{e^{x}(2 x+1)-x(x+1) e^{x}}{\left(e^{x}\right)^{2}}$

$\Rightarrow f^{\prime}(x)=\frac{(2 x+1)-x(x+1)}{e^{x}}$

$\Rightarrow f^{\prime}(x)=\frac{2 x+1-x^{2}-x}{e^{x}}$

$\Rightarrow f^{\prime}(x)=\frac{-x^{2}+x+1}{e^{x}}$

$\Rightarrow f^{\prime}(c)=\frac{-c^{2}+c+1}{e^{c}}$

 

$\therefore f^{\prime}(c)=0$

$\Rightarrow \frac{-c^{2}+c+1}{e^{c}}=0$

$\Rightarrow c^{2}-c-1=0$

$\Rightarrow c=\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}$

 

$\therefore c=\frac{1-\sqrt{5}}{2} \in(-1,0)$

Hence, the required value of $c$ is $\frac{1-\sqrt{5}}{2}$.